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:
// sync

#include <algorithm>
#include <cstdio>
#include <cstring>

const int N = 3e3 + 33, P = 1e9 + 7, div2 = (P + 1) / 2;

int n, m, x, y, a[N], b[N], dp[N][N], tmp[N][N], ans, pw2 = 1;
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", a + i), b[i] = a[i];
	std::sort(b + 1, b + n + 1);
	for (int i = 1; i <= n; i++) a[i] = std::lower_bound(b + 1, b + n + 1, a[i]) - b;
	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 <= m; i++) {
		scanf("%d%d", &x, &y);
		for (int j = 1, u, v; j <= n; j++)
			if (j != x && j != y) {
				u = 1ll * (dp[j][x] + dp[j][y]) * div2 % P;
				v = 1ll * (dp[y][j] + dp[x][j]) * div2 % P;
				dp[j][x] = u, dp[j][y] = u;
				dp[x][j] = v, dp[y][j] = v;
			}
		int u = 1ll * (dp[x][y] + dp[y][x]) * div2 % P;
		dp[x][y] = dp[y][x] = u;
		pw2 = pw2 * 2 % P;
	}
	for (int i = 1; i < n; i++)
		for (int j = i + 1; j <= n; j++) ans = (ans + dp[i][j]) % P;
	printf("%lld", 1ll * ans * pw2 % P);
}