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>
using namespace std;
typedef long long ll;

const int MOD = 1000000007;
const int MAXN = 3005;

void add(int &a, int b) { a += b; if(MOD <= a) a -= MOD; }
void div(int &a) { a = ((a&1) ? a+MOD : a) >> 1; }

int D[MAXN][MAXN], E[MAXN][MAXN];

int A[MAXN];

int N, Q, Ans;

int main() {
	ios::sync_with_stdio(false);

	cin >> N >> Q;
	for(int i = 1; i <= N; i++) cin >> A[i];

	for(int i = 1; i <= N; i++) for(int j = 1; j <= N; j++)
		D[i][j] = A[i] > A[j];
	for(int qi = Q, a, b; qi--;) {
		cin >> a >> b;
		for(int i = 1, t; i <= N; i++) {
			t = D[a][i];
			add(t, D[b][i]);
			div(t);
			E[a][i] = E[b][i] = t;

			t = D[i][a];
			add(t, D[i][b]);
			div(t);
			E[i][a] = E[i][b] = t;
		}
		{
			int t = D[a][b];
			add(t, D[b][a]);
			div(t);
			E[a][b] = E[b][a] = t;
		}
		for(int i = 1; i <= N; i++) {
			D[i][a] = E[i][a];
			D[i][b] = E[i][b];
			D[a][i] = E[a][i];
			D[b][i] = E[b][i];
			D[i][i] = 0;
		}
	}

	for(int i = 1; i < N; i++) for(int j = i+1; j <= N; j++)
		add(Ans, D[i][j]);
	for(; Q--;) {
		Ans <<= 1;
		if(MOD <= Ans) Ans -= MOD;
	}
	cout << Ans << endl;
	return 0;
}