Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

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


Output

425

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

const int N = 3000, mod = int(1e9) + 7;

int fpow(int x, int y) {
	int r = 1;
	for (; y; y >>= 1, x = 1LL * x * x % mod)
		if (y & 1) r = 1LL * r * x % mod;
	return r;
}

const int inv2 = (mod + 1) / 2;

int n, q, a[N + 5], dp[N + 5][N + 5];

int main() {
	scanf("%d%d", &n, &q);
	for (int i = 1; i <= n; ++i) scanf("%d", a + i);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j)
			dp[i][j] = a[i] > a[j];

	for (int i = 1; i <= q; ++i) {
		int x, y;
		scanf("%d%d", &x, &y);
		dp[x][y] = dp[y][x] = 1LL * (dp[x][y] + dp[y][x]) * inv2 % mod;
		for (int j = 1; j <= n; ++j)
			if (j != x && j != y) {
				dp[x][j] = dp[y][j] = 1LL * (dp[x][j] + dp[y][j]) * inv2 % mod;
				dp[j][x] = dp[j][y] = 1LL * (dp[j][x] + dp[j][y]) * inv2 % mod;
			}
	}
	int ans = 0;
	for (int i = 1; i <= n; ++i)
		for (int j = i + 1; j <= n; ++j)
			ans = (ans + dp[i][j]) % mod;
	ans = 1LL * ans * fpow(2, q) % mod;
	printf("%d\n", ans);
}