CodeForces

Discription

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you qqueries. In the i-th query you should compute and print the ugliness of a substring(continuous subsequence) of s starting at the index a_i and ending at the index b_i(inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

Output

4
3
-1

Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15

Output

-1
2
1
-1
-1

Input

Output

-1
-1

Note

In the first sample:

In the first query, ugliness("20166766") = 4 because all four sixes must be removed.
In the second query, ugliness("2016676") = 3 because all three sixes must be removed.
In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string.

In the second sample:

In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice.
In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170".

建一个节点和字符集很少的有限状态自动机，然后对于线段树每个区间用一个矩阵记录从一个状态到另一个状态的最小代价。

因为这个矩阵是满足结合律的，所以直接做就行了。

#include<bits/stdc++.h>
#define ll long long
#define lc (o<<1)
#define rc ((o<<1)|1)
#define mid (l+r>>1)
using namespace std;
const int maxn=200005;
struct node{
	int a[5][5];
	inline void init(){ memset(a,0x3f,sizeof(a));}
	node operator *(const node &u)const{
		node r; r.init();
		for(int k=0;k<5;k++)
		    for(int i=0;i<5;i++)
		        for(int j=0;j<5;j++) r.a[i][j]=min(r.a[i][j],a[i][k]+u.a[k][j]);
		return r;
	}
}S[maxn*4+5],ANS;
int n,m,le,ri;
char s[maxn];

inline void Set(node &x,int y){
	x.init(); for(int i=0;i<5;i++) x.a[i][i]=0;
	if(y==2) x.a[0][0]=1,x.a[0][1]=0;
	else if(!y) x.a[1][1]=1,x.a[1][2]=0;
	else if(y==1) x.a[2][2]=1,x.a[2][3]=0;
	else if(y==7) x.a[3][3]=1,x.a[3][4]=0;
	else if(y==6) x.a[3][3]=1,x.a[4][4]=1;
}

void build(int o,int l,int r){
	if(l==r){ Set(S[o],s[l]-'0'); return;}
	build(lc,l,mid),build(rc,mid+1,r);
	S[o]=S[lc]*S[rc];
}

void query(int o,int l,int r){
	if(l>=le&&r<=ri){ ANS=ANS*S[o]; return;}
	if(le<=mid) query(lc,l,mid);
	if(ri>mid) query(rc,mid+1,r);
}

inline void solve(){
	build(1,1,n);
	while(m--){
		ANS.init(); for(int i=0;i<5;i++) ANS.a[i][i]=0;
		scanf("%d%d",&le,&ri),query(1,1,n);
		printf("%d
",ANS.a[0][4]<=n?ANS.a[0][4]:-1);
	}
}

int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	solve();
	return 0;
}