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>
// copy ...
typedef long long ll;
const int N = 3054, mod = 1000000007, iv2 = (mod + 1) / 2;
int n, q;
int a[N], f[N][N];
ll PowerMod(ll a, int n, ll c = 1) {for (; n; n >>= 1, a = a * a % mod) if (n & 1) c = c * a % mod; return c;}
inline int half(int x) {return x = (x >> 1) + (-(x & 1) & (iv2 - mod)), x + (x >> 31 & mod);}
int main() {
	int i, j, u, v; ll I = 0;
	scanf("%d%d", &n, &q);
	for (i = 1; i <= n; ++i) scanf("%d", a + i);
	for (i = 1; i <= n; ++i)
		for (j = 1; j <= n; ++j)
			f[i][j] = a[i] > a[j];
	for (i = 0; i < q; ++i) {
		scanf("%d%d", &u, &v), f[u][v] = f[v][u] = half(f[u][v] + f[v][u]);
		for (j = 1; j <= n; ++j) if (j != u && j != v)
			f[u][j] = f[v][j] = half(f[u][j] + f[v][j]),
			f[j][u] = f[j][v] = half(f[j][u] + f[j][v]);
	}
	for (i = 1; i <= n; ++i) I += std::accumulate(f[i] + i, f[i] + (n + 1), 0ll);
	printf("%d\n", (int)PowerMod(2, q, I % mod));
	return 0;
}