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 <cstdio>
#include <algorithm>
#include <vector>
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if (ch == '-') w = -1, ch = getchar();
	while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int N(3005), Mod(1e9 + 7), inv2((Mod + 1) >> 1);
int n, Q, f[N][N], a[N], Pow = 1, ans;

int main()
{
	n = read(), Q = read();
	for (int i = 1; i <= n; i++) a[i] = read();
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) f[i][j] = a[i] < a[j];
	for (int i = 1, x, y; i <= Q; i++)
	{
		Pow = (Pow + Pow) % Mod, x = read(), y = read();
		f[x][y] = f[y][x] = 1ll * inv2 * (f[x][y] + f[y][x]) % Mod;
		for (int j = 1; j <= n; j++)
		{
			if (j == x || j == y) continue;
			f[x][j] = f[y][j] = 1ll * inv2 * (f[x][j] + f[y][j]) % Mod;
			f[j][x] = f[j][y] = 1ll * inv2 * (f[j][x] + f[j][y]) % Mod;
		}
	}
	for (int i = 1; i <= n; i++)
		for (int j = i + 1; j <= n; j++) ans = (ans + f[j][i]) % Mod;
	printf("%lld\n", 1ll * ans * Pow % Mod);
	return 0;
}