#include <bits/stdc++.h>
const int maxn = 3e5 + 10;
const int mod = 0;
using namespace std;
vector<int> q[maxn];
stack<int> S;
int link[maxn], vis[maxn], mid;
long long sum[maxn];
void dfs(int u, int pre) {
  S.push(u);
  vis[u] = 1;
  int size = q[u].size();
  for (int i = 0; i <= size - 1; i++) {
    int v = q[u][i];
    if (v != pre) {
      if (vis[v] == 0)
        dfs(v, u);
      else if (vis[v] == 1) {
        int Max = u, Min = u;
        while (!S.empty()) {
          int w = S.top();
          S.pop();
          Max = max(Max, w);
          Min = min(Min, w);
          if (w == v) break;
        }
        link[Min] = Max;
      }
    }
  }
  if (!S.empty() && S.top() == u) S.pop();
  vis[u] = 2;
}
int main() {
  int n, m, a, b, k, l, r, ans;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    q[a].push_back(b);
    q[b].push_back(a);
  }
  for (int i = 1; i <= n + 1; i++) link[i] = n + 1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, -1);
  sum[n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    link[i] = min(link[i], link[i + 1]);
    sum[i] = link[i] - i + sum[i + 1];
  }
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d%d", &l, &r);
    int L = l, R = r, p = l;
    while (L <= R) {
      mid = L + R >> 1;
      if (link[mid] <= r)
        L = mid + 1;
      else {
        p = mid;
        R = mid - 1;
      }
    }
    long long ans = sum[l] - sum[p] + 1LL * (r - p + 2) * (r - p + 1) / 2;
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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>
const int maxn = 3e5 + 10;
const int mod = 0;
using namespace std;
vector<int> q[maxn];
stack<int> S;
int link[maxn], vis[maxn], mid;
long long sum[maxn];
void dfs(int u, int pre) {
  S.push(u);
  vis[u] = 1;
  int size = q[u].size();
  for (int i = 0; i <= size - 1; i++) {
    int v = q[u][i];
    if (v != pre) {
      if (vis[v] == 0)
        dfs(v, u);
      else if (vis[v] == 1) {
        int Max = u, Min = u;
        while (!S.empty()) {
          int w = S.top();
          S.pop();
          Max = max(Max, w);
          Min = min(Min, w);
          if (w == v) break;
        }
        link[Min] = Max;
      }
    }
  }
  if (!S.empty() && S.top() == u) S.pop();
  vis[u] = 2;
}
int main() {
  int n, m, a, b, k, l, r, ans;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    q[a].push_back(b);
    q[b].push_back(a);
  }
  for (int i = 1; i <= n + 1; i++) link[i] = n + 1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, -1);
  sum[n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    link[i] = min(link[i], link[i + 1]);
    sum[i] = link[i] - i + sum[i + 1];
  }
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d%d", &l, &r);
    int L = l, R = r, p = l;
    while (L <= R) {
      mid = L + R >> 1;
      if (link[mid] <= r)
        L = mid + 1;
      else {
        p = mid;
        R = mid - 1;
      }
    }
    long long ans = sum[l] - sum[p] + 1LL * (r - p + 2) * (r - p + 1) / 2;
    printf("%I64d\n", ans);
  }
  return 0;
}

```