codeforces 702A A. Maximum Increase(水题)

题目链接：

You are given array consisting of n integers. Your task is to find the maximum length of an increasing subarray of the given array.

A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.

Input

The first line contains single positive integer n (1 ≤ n ≤ 105) — the number of integers.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the maximum length of an increasing subarray of the given array.

Examples

input

5
1 7 2 11 15

output

input

6
100 100 100 100 100 100

output

input

3
1 2 3

output

3

题意：

求最长的上升子串;

思路：

水题;

AC代码：

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┆　　┃　  　┃　　┆　　　　　　 
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┆　　┃　 AC代马 　　┣┓┆ 
┆　　┃　        　　┏┛┆ 
┆　　┗┓┓┏━┳┓┏┛ ┆ 
┆　　　┃┫┫　┃┫┫ ┆ 
┆　　　┗┻┛　┗┻┛ ┆       
************************************************ */  


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=(1<<8);
const double eps=1e-8;

int a[N],g[N],d[N];
int main()
{       
        int n;
        read(n);
        For(i,1,n)read(a[i]);
        int ans=1,temp=1;
        For(i,2,n)
        {
            if(a[i]>a[i-1])temp++;
            else 
            {
                ans=max(ans,temp);
                temp=1;
            }
        }
        ans=max(ans,temp);
        /*
        For(i,1,n)
        {
            int k=lower_bound(g+1,g+n+1,a[i])-g;
            d[i]=k;
            g[k]=a[i];
        }
        cout<<d[n]<<endl;*/
        cout<<ans<<endl;
        
        return 0;
}