#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n, m, q, low[N], vis[N];
long long pre[N];
vector<int> es[N];
stack<int> stk;
void dfs(int u, int par) {
  stk.push(u);
  vis[u] = 1;
  for (auto v : es[u])
    if (v != par) {
      if (!vis[v])
        dfs(v, u);
      else if (vis[v] == 1) {
        int mn = v, mx = v;
        while (!stk.empty()) {
          int cur = stk.top();
          stk.pop();
          mn = min(mn, cur);
          mx = max(mx, cur);
          if (cur == v) break;
        }
        low[mx] = mn;
      }
    }
  if (!stk.empty() && stk.top() == u) stk.pop();
  vis[u] = 2;
}
void solve() {
  scanf("%d%d", &n, &m);
  for (int _ = 0; _ < m; _++) {
    int a, b;
    scanf("%d%d", &a, &b);
    es[a].push_back(b);
    es[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  for (int i = 1; i <= n; i++) {
    low[i] = max(low[i], low[i - 1]);
    pre[i] = pre[i - 1] + i - low[i];
  }
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int lo = l, hi = r;
    while (lo < hi) {
      int mid = (lo + hi + 1) / 2;
      if (low[mid] < l)
        lo = mid;
      else
        hi = mid - 1;
    }
    printf("%lld\n", pre[r] - pre[lo] + 1ll * (lo - l + 1) * (lo - l + 2) / 2);
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int N = 3e5 + 5;
int n, m, q, low[N], vis[N];
long long pre[N];
vector<int> es[N];
stack<int> stk;
void dfs(int u, int par) {
  stk.push(u);
  vis[u] = 1;
  for (auto v : es[u])
    if (v != par) {
      if (!vis[v])
        dfs(v, u);
      else if (vis[v] == 1) {
        int mn = v, mx = v;
        while (!stk.empty()) {
          int cur = stk.top();
          stk.pop();
          mn = min(mn, cur);
          mx = max(mx, cur);
          if (cur == v) break;
        }
        low[mx] = mn;
      }
    }
  if (!stk.empty() && stk.top() == u) stk.pop();
  vis[u] = 2;
}
void solve() {
  scanf("%d%d", &n, &m);
  for (int _ = 0; _ < m; _++) {
    int a, b;
    scanf("%d%d", &a, &b);
    es[a].push_back(b);
    es[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  for (int i = 1; i <= n; i++) {
    low[i] = max(low[i], low[i - 1]);
    pre[i] = pre[i - 1] + i - low[i];
  }
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int lo = l, hi = r;
    while (lo < hi) {
      int mid = (lo + hi + 1) / 2;
      if (low[mid] < l)
        lo = mid;
      else
        hi = mid - 1;
    }
    printf("%lld\n", pre[r] - pre[lo] + 1ll * (lo - l + 1) * (lo - l + 2) / 2);
  }
}
int main() {
  solve();
  return 0;
}

```