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 MAXN = 1e6 + 100;
int dfn[MAXN] = {0}, vis[MAXN] = {0}, low[MAXN] = {0}, tot = 0,
    post[MAXN] = {0};
long long sum[MAXN];
stack<int> S;
vector<int> E[MAXN];
void tarjan(int par, int x) {
  low[x] = dfn[x] = ++tot;
  vis[x] = 1;
  S.push(x);
  for (int i = 0; i < E[x].size(); i++) {
    int v = E[x][i];
    if (v == par) continue;
    if (!dfn[v]) {
      tarjan(x, v);
      low[x] = min(low[x], low[v]);
    } else if (vis[v])
      low[x] = min(low[x], dfn[v]);
  }
  if (low[x] == dfn[x]) {
    int mi = x, ma = x;
    while (1) {
      int now = S.top();
      S.pop();
      vis[now] = 0;
      mi = min(mi, now);
      ma = max(ma, now);
      if (now == x) break;
    }
    if (mi != ma) post[mi] = ma - 1;
  }
}
int main() {
  int n, m, u, v, l, r;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  memset(post, 0x3f, sizeof(post));
  for (int i = 1; i <= n; i++)
    if (dfn[i] == 0) tarjan(0, i);
  post[n] = n;
  for (int i = n - 1; i >= 1; i--) post[i] = min(post[i], post[i + 1]);
  sum[0] = 0;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + post[i];
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &l, &r);
    int p = upper_bound(post + l, post + r, r) - post;
    long long ans = sum[p - 1] - sum[l - 1] + 1LL * r * (r - p + 1) -
                    1LL * (l + r - 2) * (r - l + 1) / 2;
    printf("%lld\n", ans);
  }
}
