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;
inline int read() {
  int X = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') X = X * 10 + c - '0', c = getchar();
  return X * w;
}
const int N = 300000 + 10, M = 300000 + 10;
int n, m, q;
int nxt[N];
long long sum[N];
struct edge {
  int v, nxt;
} e[M << 1];
int head[N];
inline void addEdge(int u, int v) {
  static int c = 0;
  e[++c] = (edge){v, head[u]}, head[u] = c;
}
int vis[N], in[N], sta[N], top = 0;
inline void dfs(int u, int fa) {
  vis[u] = 1, sta[++top] = u;
  for (register int 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;
      while ("longge ak ioi") {
        int x = sta[top--];
        in[x] = 1;
        mx = max(mx, x), mn = min(mn, x);
        if (x == v) break;
      }
      nxt[mn] = min(nxt[mn], mx - 1);
    }
  }
  if (!in[u]) --top;
}
int main() {
  n = read(), m = read();
  for (register int i = 1; i <= m; ++i) {
    int u = read(), v = read();
    addEdge(u, v), addEdge(v, u);
  }
  for (register int i = 1; i <= n; ++i) nxt[i] = n;
  for (register int i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i, 0);
  for (register int i = n - 1; i; --i) nxt[i] = min(nxt[i + 1], nxt[i]);
  for (register int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + nxt[i];
  q = read();
  while (q--) {
    int l = read(), r = read();
    int L = l, R = r;
    while (L < R) {
      int mid = (L + R + 1) >> 1;
      if (nxt[mid] < r)
        L = mid;
      else
        R = mid - 1;
    }
    if (nxt[L] > r) --L;
    printf("%lld\n", 1ll * (r - L) * (r - L + 1) / 2 + sum[L] - sum[l - 1] -
                         1ll * (l + L - 2) * (L - l + 1) / 2);
  }
  return 0;
}
