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 = 300005;
struct sl {
  int to, last;
} line[2 * N];
int n, m, z, x, y, Q, dfn[N], low[N], tot, last[N], num, l, r;
long long a[N], sum[N];
stack<int> sta;
bool c[N], cc[N];
void add(int x, int y) {
  line[++num].to = y;
  line[num].last = last[x];
  last[x] = num;
}
void dfs(int l, int fa) {
  sta.push(l);
  dfn[l] = low[l] = ++tot;
  c[l] = cc[l] = true;
  for (int j = last[l]; j; j = line[j].last)
    if (line[j].to != fa) {
      if (c[line[j].to] == false) {
        dfs(line[j].to, l);
        low[l] = min(low[l], low[line[j].to]);
      } else if (cc[line[j].to] == true)
        low[l] = min(low[l], dfn[line[j].to]);
    }
  if (low[l] == dfn[l]) {
    if (sta.top() != l) {
      int Mx = 0, Mn = 100000000;
      while (sta.top() != l) {
        Mx = max(Mx, sta.top());
        Mn = min(Mn, sta.top());
        cc[sta.top()] = false;
        sta.pop();
      }
      Mx = max(Mx, sta.top());
      Mn = min(Mn, sta.top());
      cc[sta.top()] = false;
      sta.pop();
      a[Mn] = Mx;
    } else
      sta.pop();
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  for (int i = 1; i <= n + 1; ++i) a[i] = n + 1;
  for (int i = 1; i <= n; ++i)
    if (!c[i]) dfs(i, 0);
  for (int i = n; i >= 1; --i) {
    a[i] = min(a[i], a[i + 1]);
    sum[i] = sum[i + 1] + a[i] - i;
  }
  scanf("%d", &Q);
  for (int i = 1; i <= Q; ++i) {
    scanf("%d%d", &l, &r);
    x = lower_bound(a + l, a + r + 1, r + 1) - a;
    printf("%I64d\n", 1LL * (r - x + 1) * (r - x + 2) / 2 + sum[l] - sum[x]);
  }
}
