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;
typedef int arr[600010];
arr head, pnt, Next, dfn, low, vis, flag, st, p;
long long s[300010];
int tot, n, m, q, tim, tp;
inline void add(int a, int b) {
  Next[++tot] = head[a];
  head[a] = tot;
  pnt[tot] = b;
  Next[++tot] = head[b];
  head[b] = tot;
  pnt[tot] = a;
}
inline void dfs(int x, int fa) {
  low[x] = dfn[x] = ++tim;
  vis[x] = 1;
  st[++tp] = x;
  flag[x] = 1;
  for (int i = head[x]; i; i = Next[i])
    if (pnt[i] != fa) {
      if (!vis[pnt[i]]) {
        dfs(pnt[i], x);
        low[x] = min(low[x], low[pnt[i]]);
      } else if (flag[pnt[i]])
        low[x] = min(low[x], dfn[pnt[i]]);
    }
  if (low[x] == dfn[x]) {
    int tt, Mn = INT_MAX, Mx = 0;
    do {
      tt = st[tp--];
      flag[tt] = 0;
      Mn = min(Mn, tt);
      Mx = max(Mx, tt);
    } while (tt != x);
    if (Mx != Mn) p[Mx] = max(p[Mx], Mn);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, a, b; i <= m; i++) scanf("%d%d", &a, &b), add(a, b);
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) dfs(i, 0);
  for (int i = 1; i <= n; i++)
    p[i] = max(p[i], p[i - 1]), s[i] = s[i - 1] + p[i];
  scanf("%d", &q);
  for (int i = 1, l, r, x, y; i <= q; i++) {
    scanf("%d%d", &x, &y);
    l = x, r = y;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (p[mid] >= x)
        r = mid - 1;
      else
        l = mid + 1;
    }
    printf("%I64d\n", 1LL * (x + y) * (y - x + 1) / 2 + s[l - 1] - s[y] -
                          1LL * (x - 1) * (l - x));
  }
  return 0;
}
