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 mod = 1000000007;
typedef long long LL;
void reduce(int & x) { x += x >> 31 & mod; }
int mul(int a, int b) { return (LL) a * b % mod; }
int pow(int a, int b, int res = 1) {
	for (; b; b >>= 1, a = mul(a, a)) if (b & 1) res = mul(res, a);
	return res;
}
const int MAXN = 3010;
int A[MAXN], n, Q;
int f[MAXN][MAXN];
void eff(int & x, int & y) {
	x = y = mul(x + y, mod + 1 >> 1);
}
int main() {
	std::ios_base::sync_with_stdio(false), 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 = i + 1; j <= n; ++j) if (A[i] != A[j])
			A[i] < A[j] ? ++f[i][j] : ++f[j][i];
	for (int i = 1; i <= Q; ++i) {
		int x, y; std::cin >> x >> y;
		for (int j = 1; j <= n; ++j) if (j != x && j != y) {
			eff(f[x][j], f[y][j]);
			eff(f[j][x], f[j][y]);
		}
		eff(f[x][y], f[y][x]);
	}
	int ans = 0;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j < i; ++j)
			reduce(ans += f[i][j] - mod);
	ans = pow(2, Q, ans);
	std::cout << ans << std::endl;
	return 0;
}