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 <iostream>

using i64 = long long;

const int N = 3000 + 7;
const i64 MOD = 1000000007;
const i64 Inv2 = MOD - MOD / 2;

int n, q;
int a[N];
i64 f[N][N], g[N][N], ans, coe = 1;

int main() {
//	freopen("code.in", "r", stdin);
//	freopen("code.out", "w", stdout);

	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) f[i][j] = g[i][j] = a[i] > a[j];
	for (int x, y; q; --q) {
		scanf("%d%d", &x, &y);
		for (int i = 1; i <= n; ++i) {
			if (i != x) g[i][x] = (f[i][x] + f[i][y]) * Inv2 % MOD, g[x][i] = (f[x][i] + f[y][i]) * Inv2 % MOD;
			if (i != y) g[i][y] = (f[i][x] + f[i][y]) * Inv2 % MOD, g[y][i] = (f[x][i] + f[y][i]) * Inv2 % MOD;
		}
		g[x][y] = g[y][x] = (f[x][y] + f[y][x]) * Inv2 % MOD;
		for (int i = 1; i <= n; ++i)
			f[i][x] = g[i][x], f[x][i] = g[x][i], f[i][y] = g[i][y], f[y][i] = g[y][i];
		f[x][y] = f[y][x] = g[x][y];
		coe = coe * 2 % MOD;
	}
	for (int i = 1; i <= n; ++i)
		for (int j = i + 1; j <= n; ++j) ( ans += f[i][j] ) %= MOD;
	printf("%lld ", ans * coe % MOD);
	return 0;
}