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;
struct edge {
  int v, nxt;
} e[600005];
int n, m, q, c, top, nxt[300005], head[300005], vis[300005], in[300005],
    sta[300005];
long long sum[300005];
void add(int u, int v) {
  e[++c] = (edge){v, head[u]};
  head[u] = c;
}
void dfs(int u, int fa) {
  int i, v, x;
  vis[u] = 1;
  sta[++top] = u;
  for (i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    if (!vis[v])
      dfs(v, u);
    else if (!in[v]) {
      int mx = 0, mn = n + 1;
      do {
        x = sta[top--];
        in[x] = 1;
        mx = max(mx, x);
        mn = min(mn, x);
      } while (x != v);
      nxt[mn] = min(nxt[mn], mx - 1);
    }
  }
  if (!in[u]) top--;
}
int main() {
  int i, u, v, l, r, l1, r1, mid;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  for (i = 1; i <= n; i++) nxt[i] = n;
  for (i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  for (i = n - 1; i; i--) nxt[i] = min(nxt[i + 1], nxt[i]);
  for (i = 1; i <= n; i++) sum[i] = sum[i - 1] + nxt[i];
  cin >> q;
  while (q--) {
    scanf("%d%d", &l, &r);
    l1 = l;
    r1 = r;
    while (l1 < r1) {
      mid = (l1 + r1 + 1) / 2;
      if (nxt[mid] < r)
        l1 = mid;
      else
        r1 = mid - 1;
    }
    if (nxt[l1] > r) l1--;
    cout << 1ll * (r - l1) * (r - l1 + 1) / 2 + sum[l1] - sum[l - 1] -
                1ll * (l + l1 - 2) * (l1 - l + 1) / 2
         << endl;
  }
  return 0;
}
