#include <bits/stdc++.h>
using namespace std;
vector<int> graph[300030];
long long to[300030], sum[300030];
int vis[300030], dep[300030], parent[300030], group[300030];
int cyc, cnt[300030], gcnt[300030];
void dfs(int node, int p, int d = 1) {
  if (vis[node]) return;
  vis[node] = 1;
  dep[node] = d;
  parent[node] = p;
  for (int to : graph[node]) {
    if (vis[to] && dep[to] < d && to != p) {
      ++cyc;
      for (int k = node; k > 0; k = parent[k]) {
        group[k] = cyc;
        gcnt[cyc]++;
        if (k == to) break;
      }
    }
    dfs(to, node, d + 1);
  }
}
void count(int x, int val) {
  if (group[x] > 0) cnt[group[x]] += val;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, -1);
  }
  int e = 1;
  count(e, 1);
  for (int i = 1; i <= n; i++) {
    to[i] = ((e) < (n) ? (e) : (n));
    while (++e <= n) {
      count(e, 1);
      if (group[e] > 0 && cnt[group[e]] == gcnt[group[e]]) {
        count(e, -1);
        --e;
        break;
      }
      to[i] = e;
    }
    sum[i] = sum[i - 1] + to[i];
    count(i, -1);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    long long l, r, pos = x - 1;
    l = x;
    r = y;
    while (l <= r) {
      long long m = (l + r) >> 1;
      if (to[m] <= y) {
        pos = m;
        l = m + 1;
      } else {
        r = m - 1;
      }
    }
    long long l1 = pos - x + 1;
    long long l2 = y - pos;
    printf("%lld\n", sum[pos] - sum[x - 1] - l1 * (l1 - 1) / 2 - (x - 1) * l1 +
                         l2 * (l2 + 1) / 2);
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> graph[300030];
long long to[300030], sum[300030];
int vis[300030], dep[300030], parent[300030], group[300030];
int cyc, cnt[300030], gcnt[300030];
void dfs(int node, int p, int d = 1) {
  if (vis[node]) return;
  vis[node] = 1;
  dep[node] = d;
  parent[node] = p;
  for (int to : graph[node]) {
    if (vis[to] && dep[to] < d && to != p) {
      ++cyc;
      for (int k = node; k > 0; k = parent[k]) {
        group[k] = cyc;
        gcnt[cyc]++;
        if (k == to) break;
      }
    }
    dfs(to, node, d + 1);
  }
}
void count(int x, int val) {
  if (group[x] > 0) cnt[group[x]] += val;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, -1);
  }
  int e = 1;
  count(e, 1);
  for (int i = 1; i <= n; i++) {
    to[i] = ((e) < (n) ? (e) : (n));
    while (++e <= n) {
      count(e, 1);
      if (group[e] > 0 && cnt[group[e]] == gcnt[group[e]]) {
        count(e, -1);
        --e;
        break;
      }
      to[i] = e;
    }
    sum[i] = sum[i - 1] + to[i];
    count(i, -1);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    long long l, r, pos = x - 1;
    l = x;
    r = y;
    while (l <= r) {
      long long m = (l + r) >> 1;
      if (to[m] <= y) {
        pos = m;
        l = m + 1;
      } else {
        r = m - 1;
      }
    }
    long long l1 = pos - x + 1;
    long long l2 = y - pos;
    printf("%lld\n", sum[pos] - sum[x - 1] - l1 * (l1 - 1) / 2 - (x - 1) * l1 +
                         l2 * (l2 + 1) / 2);
  }
  return 0;
}

```