You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n, m, q;
int p[N], st[N], dp[N];
long long sum[N];
vector<int> g[N];
int vis[N];
void cycle(int a, int x) {
  int l = n + 2, r = -1;
  l = min(l, x);
  r = max(r, x);
  while (x != a) {
    x = p[x];
    l = min(l, x);
    r = max(r, x);
  }
  st[l] = r;
}
void dfs(int x) {
  vis[x] = 1;
  for (auto it : g[x]) {
    if (!vis[it]) {
      p[it] = x;
      dfs(it);
    } else if (it != p[x] and vis[it] == 1) {
      cycle(it, x);
    }
  }
  vis[x] = 2;
}
int main() {
  scanf("%d%d", &n, &m);
  int u, v;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i);
    dp[i] = n;
  }
  dp[n] = n;
  for (int i = n - 1; i > 0; i--) {
    dp[i] = dp[i + 1];
    if (st[i] > 0) dp[i] = min(dp[i], st[i] - 1);
  }
  for (int i = 1; i <= n; i++) {
    dp[i] -= i;
    sum[i] = sum[i - 1] + dp[i];
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    int l = a, r = b;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (dp[mid] + mid >= b)
        r = mid;
      else
        l = mid + 1;
    }
    long long ans = 0;
    ans += (1ll * (b - l + 1) * (b - l + 2)) / 2;
    ans += sum[l - 1] - sum[a - 1] + l - a;
    printf("%lld\n", ans);
  }
}
