coderforce Educational Codeforces Round 10 C. Foe Pairs（贪心）

C. Foe Pairs

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a permutation p of length n. Also you are given m foe pairs (a_i, b_i) (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i).

Your task is to count the number of different intervals (x, y) (1 ≤ x ≤ y ≤ n) that do not contain any foe pairs. So you shouldn't count intervals (x, y) that contain at least one foe pair in it (the positions and order of the values from the foe pair are not important).

Consider some example: p = [1, 3, 2, 4] and foe pairs are {(3, 2), (4, 2)}. The interval (1, 3) is incorrect because it contains a foe pair (3, 2). The interval (1, 4) is also incorrect because it contains two foe pairs (3, 2) and (4, 2). But the interval (1, 2) is correct because it doesn't contain any foe pair.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 3·10⁵) — the length of the permutation p and the number of foe pairs.

The second line contains n distinct integers p_i (1 ≤ p_i ≤ n) — the elements of the permutation p.

Each of the next m lines contains two integers (a_i, b_i) (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the i-th foe pair. Note a foe pair can appear multiple times in the given list.

Output

Print the only integer c — the number of different intervals (x, y) that does not contain any foe pairs.

Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Examples

Input

4 2
1 3 2 4
3 2
2 4

Output

5

Input

9 5
9 7 2 3 1 4 6 5 8
1 6
4 5
2 7
7 2
2 7

Output

20

输入：n，m。n代表输入数组范围，m代表pair的组数。下面1行有n个，表示排列中的数。接下来m行输入foe的值

输出：输出不包含foe组合的区间总数，可能会超int。

解法：对于输入的数组，我们可以用pos[i]来保存i的位置，对于任意一个位置，我们可以定义mark[i]为以位置i结尾的不能取到的区间数。根据这个我们可以推出mark[i]=max(mark[i],l),l表示foe的左边界。然后如果没有不可以取得pair的话，以i结尾的能取的组数为i。所以答案的总数为：i-mark[i]的总和。

AC：

#include <algorithm>
#include <string>
#include <iostream>
#include <string.h>
#include<stdio.h>
using namespace std;
#define maxn 1000100
#define ll __int64
int mark[maxn],pos[maxn];
int main()
{
    int n,m,v,x,y,l,r;
    ll ans=0;
    cin>>n>>m;
    memset(mark,0,sizeof(mark));
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&v);
        pos[v]=i;
    }
    while(m--)
    {


        scanf("%d%d",&x,&y);
        l=pos[x],r=pos[y];
        if(l>r) swap(l,r);
        mark[r]=max(mark[r],l);
    }
    mark[0]=0;
    for(int i=1;i<=n;i++)
    {
        mark[i]=max(mark[i],mark[i-1]);
        ans+=(i-mark[i]);
    }
    cout<<ans<<endl;


}