CF652 C ma~ 对我来说只要是牵扯到方法的题目都是好题

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 (ai, bi) (1 ≤ ai, bi ≤ n, ai ≠ bi).

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 pi (1 ≤ pi ≤ n) — the elements of the permutation p.

Each of the next m lines contains two integers (ai, bi) (1 ≤ ai, bi ≤ n, ai ≠ bi) — 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 21 3 2 43 22 4

output

5

input

9 59 7 2 3 1 4 6 5 81 64 52 77 22 7

output

20

Note

In the first example the intervals from the answer are (1, 1), (1, 2), (2, 2), (3, 3) and (4, 4).

其实这道题目我想了好久。。。感觉果然有点笨呐，竟然没有想到。。。（可能是因为一直想着别人的题解忘了自己应该怎么去思考了吧。呜~多思考哟）（好吧，其实是自己理解错误了所以最后半天没有写出来TAT）

恩，题目大意就是给你一些数字，然后让你找到符合条件的区间数有几个。这个区间中不能有敌人的pair。

思路：枚举，从右开始枚举，右端枚举该数组到左端即可。

用pos数组表示当前坐标所在的位置。（注意，这个区间是题目所给出的数列的区间）

然后就是用r数组来保存这些位置的右端，其中r[i]中，i是表示当前的左端点（然后要从右边开始枚举，这样就不需要考虑左端点了，只需要知道右端点的最小值即可）。然后r数组的值就是敌人的位置的（也就是有区间）。然后这个时候只需要用rb储存又断点就可以了。

#include<bits stdc="" h="">using namespace std;typedef long long ll;const int maxn = 3 * 100000 + 5;int n, m;int pos[maxn];int r[maxn];void init(){ for (int i = 1; i <= n; i++){ int x; scanf("%d", &x); pos[x] = i; r[i] = n + 1; } for (int i = 1; i <= m; i++){ int a, b; scanf("%d%d", &a, &b); a = pos[a]; b = pos[b]; if (a > b) swap(a, b); r[a] = min(r[a], b); } int rb = n + 1; ll res = 0; for (int i = n; i > 0; i--){ rb = min(rb, r[i]); res += rb - i; } printf("%I64d\n", res);}int main(){ while (scanf("%d%d", &n, &m) == 2){ init(); } return 0;}</bits>