Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char ch = getchar();
  int nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  int ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
int b[5005];
int n, q;
struct Node {
  vector<int> v;
  map<pair<int, int>, int> m;
};
Node a[5005 * 4];
vector<pair<int, int> > c;
int ans[5005 * 20];
void build(int u, int l, int r) {
  if (l == r) {
    a[u].v.push_back(b[l]);
    return;
  }
  int mid = (l + r) / 2;
  build((u << 1), l, mid);
  build((u << 1 | 1), mid + 1, r);
  int i = 0, j = 0;
  while (i < (int)a[(u << 1)].v.size() && j < (int)a[(u << 1 | 1)].v.size()) {
    if (a[(u << 1)].v[i] < a[(u << 1 | 1)].v[j])
      a[u].v.push_back(a[(u << 1)].v[i++]);
    else
      a[u].v.push_back(a[(u << 1 | 1)].v[j++]);
  }
  while (i < (int)a[(u << 1)].v.size()) a[u].v.push_back(a[(u << 1)].v[i++]);
  while (j < (int)a[(u << 1 | 1)].v.size())
    a[u].v.push_back(a[(u << 1 | 1)].v[j++]);
}
int solve(int u, int L, int R) {
  int l = lower_bound(a[u].v.begin(), a[u].v.end(), L) - a[u].v.begin();
  int r = upper_bound(a[u].v.begin(), a[u].v.end(), R) - a[u].v.begin();
  if (l == (int)a[u].v.size() || l == r) return -1;
  if (l + 1 == r) return a[u].v[l];
  if (a[u].m[make_pair(l, r)]) return a[u].m[make_pair(l, r)];
  int x = solve((u << 1), L, R), y = solve((u << 1 | 1), L, R);
  if (x != -1 && y != -1) {
    c.emplace_back(x, y);
    return a[u].m[make_pair(l, r)] = c.size() + n;
  } else
    return a[u].m[make_pair(l, r)] = max(x, y);
}
signed main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) b[read()] = i;
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int l = read(), r = read();
    ans[i] = solve(1, l, r);
  }
  cout << c.size() + n << endl;
  for (auto [x, y] : c) printf("%d %d\n", x, y);
  for (int i = 1; i <= q; i++) printf("%d ", ans[i]);
  return 0;
}