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 _ = 3e5 + 10;
const int INF = 0x3f3f3f3f;
int N, M, R[_];
vector<int> E[_];
int tot, head[_], to[_ << 1], nxt[_ << 1];
long long sum[_];
inline int ty() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void adde(int x, int y) {
  to[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
int num, dfn[_], low[_];
int stk[_], top;
void tarjan(int x, const int root) {
  dfn[x] = low[x] = ++num;
  stk[++top] = x;
  if (x == root && head[x] == 0) return;
  for (int i = head[x]; i; i = nxt[i]) {
    int y = to[i];
    if (!dfn[y]) {
      tarjan(y, root);
      low[x] = min(low[x], low[y]);
      if (dfn[x] <= low[y]) {
        int z, maxx = 0, minx = INF;
        int cnt = 0;
        do {
          ++cnt;
          z = stk[top--];
          maxx = max(maxx, z);
          minx = min(minx, z);
        } while (z != y);
        ++cnt;
        maxx = max(maxx, x);
        minx = min(minx, x);
        if (cnt >= 3) R[minx] = min(R[minx], maxx);
      }
    } else
      low[x] = min(low[x], dfn[y]);
  }
}
int main() {
  N = ty(), M = ty();
  int T, x, y;
  for (int i = 1; i <= M; ++i) {
    x = ty(), y = ty();
    adde(x, y);
    adde(y, x);
  }
  fill(R + 1, R + N + 1, INF);
  for (int i = 1; i <= N; ++i)
    if (!dfn[i]) tarjan(i, i);
  for (int i = N - 1; i >= 1; --i) R[i] = min(R[i], R[i + 1]);
  for (int i = 1; i <= N; ++i) sum[i] = sum[i - 1] + R[i];
  long long ans;
  T = ty();
  while (T--) {
    x = ty(), y = ty();
    int l = x, r = y;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (R[mid] > y)
        r = mid;
      else
        l = mid + 1;
    }
    ans = 0;
    ans += sum[l - 1] - sum[x - 1] - 1ll * (x + l - 1) * (l - x) / 2;
    int len = y - l + 1;
    ans += 1ll * len * (len + 1) / 2;
    printf("%lld\n", ans);
  }
  return 0;
}
