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>

const int N = 3005, mod = 1000000007, i2 = mod + 1 >> 1;
typedef long long LL;

int n, q, dp[N][N], tmp[N][N], x, y, A[N];

void copy(int x, int y) { tmp[x][y] = dp[x][y]; }
void update(int X, int Y) {
	int tx = X, ty = Y;
	if (tx == x) tx = y; else if (tx == y) tx = x;
	if (ty == x) ty = y; else if (ty == y) ty = x;
	dp[X][Y] = (LL) i2 * (tmp[X][Y] + tmp[tx][ty]) % mod;
}

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n >> q;
	for (int i = 1; i <= n; ++i) std::cin >> 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) {
		std::cin >> x >> y;
		for (int j = 1; j <= n; ++j)
			copy(x, j), copy(j, x), copy(y, j), copy(j, y);
		for (int j = 1; j <= n; ++j)
			update(x, j), update(j, x), update(y, j), update(j, y);
	}
	int ans = 0;
	for (int i = 1; i <= n; ++i)
		for (int j = i + 1; j <= n; ++j)
			ans += dp[i][j] - mod, ans += ans >> 31 & mod;
	for (int i = 1; i <= q; ++i)
		ans += ans - mod, ans += ans >> 31 & mod;
	std::cout << ans << '\n';
	return 0;
}