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 = 600010, inf = N;
int n, m;
int head[N], cnt = 1;
struct edge {
  int to, next;
} e[N << 1];
void adde(int a, int b) {
  e[++cnt].to = b;
  e[cnt].next = head[a];
  head[a] = cnt;
}
int nxt[N], st[N], vis[N], top, mn, mx, in[N];
void dfs(int k, int fa) {
  vis[k] = 1;
  st[++top] = k;
  for (int i = head[k]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    if (!vis[v])
      dfs(v, k);
    else if (!in[v]) {
      mx = 0;
      mn = inf;
      while (top) {
        mx = max(st[top], mx);
        mn = min(mn, st[top]);
        in[st[top]] = 1;
        top--;
        if (st[top + 1] == v) break;
      }
      nxt[mn] = min(nxt[mn], mx);
    }
  }
  if (!in[k]) top--;
}
long long sum[N];
int main() {
  scanf("%d%d", &n, &m);
  int u, v, q;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u, &v);
    adde(u, v);
    adde(v, u);
  }
  for (int i = 1; i <= n + 1; i++) nxt[i] = n + 1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  for (int i = n; i; i--) nxt[i] = min(nxt[i + 1], nxt[i]);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + nxt[i];
  scanf("%d", &q);
  int L, R;
  while (q--) {
    scanf("%d%d", &L, &R);
    int l = L, r = R, res = L - 1;
    while (l <= r) {
      int mid = l + r >> 1;
      if (nxt[mid] > R)
        r = mid - 1;
      else
        l = mid + 1, res = mid;
    }
    printf("%lld\n", 1ll * (R - res) * (R - res + 1) / 2 + sum[res] -
                         sum[L - 1] - 1ll * (res + L) * (res - L + 1) / 2);
  }
  return 0;
}
