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;
const int MAXN = (1 << 12) + 5, Sz = (1 << 4) + 5, B = (1 << 8),
          MAXM = 2.2e6 + 5;
char buf[1 << 12], *pp1 = buf, *pp2 = buf, nc;
int ny;
inline char gc() {
  return pp1 == pp2 &&
                 (pp2 = (pp1 = buf) + fread(buf, 1, 1 << 12, stdin), pp1 == pp2)
             ? EOF
             : *pp1++;
}
inline int read() {
  int x = 0;
  for (ny = 1; nc = gc(), (nc < 48 || nc > 57) && nc != EOF;)
    if (nc == 45) ny = -1;
  if (nc < 0) return nc;
  for (x = nc - 48; nc = gc(), 47 < nc && nc < 58 && nc != EOF;
       x = (x << 3) + (x << 1) + (nc ^ 48))
    ;
  return x * ny;
}
int n, m, a[MAXN], pos[MAXN], cnt;
pair<int, int> ans[MAXM];
inline int Merge(int x, int y) { return ans[++cnt] = make_pair(x, y), cnt; }
struct Block {
  int L, R, pos[B + 5], rev[MAXN], N, res[B + 5][B + 5], preL[B + 5],
      preR[B + 5], nxtL[B + 5], nxtR[B + 5];
  inline void solve(int l, int r, vector<int> p) {
    if (l == r) {
      res[p[0]][p[0]] = pos[p[0]];
      return;
    }
    int mid = l + r >> 1;
    vector<int> u(0), v(0);
    for (auto o : p)
      if (a[pos[o]] <= L + mid - 1)
        u.push_back(o);
      else
        v.push_back(o);
    solve(l, mid, u), solve(mid + 1, r, v);
    int t = 0;
    for (auto i : p) {
      if (a[pos[i]] <= L + mid - 1)
        preL[t] = t, preR[t] = t ? preR[t - 1] : -1;
      else
        preR[t] = t, preL[t] = t ? preL[t - 1] : -1;
      t++;
    }
    t = p.size() - 1, reverse(p.begin(), p.end());
    for (auto i : p) {
      if (a[pos[i]] <= L + mid - 1)
        nxtL[t] = t, nxtR[t] = (t == p.size() - 1 ? p.size() : nxtR[t + 1]);
      else
        nxtL[t] = (t == p.size() - 1 ? p.size() : nxtL[t + 1]), nxtR[t] = t;
      t--;
    }
    reverse(p.begin(), p.end());
    static int tmp[B + 5][B + 5];
    for (int i = 0; i < p.size(); i++)
      for (int j = i; j < p.size(); j++) tmp[p[i]][p[j]] = res[p[i]][p[j]];
    for (int i = 0; i < p.size(); i++)
      for (int j = i; j < p.size(); j++) {
        int x, y;
        if (a[pos[p[i]]] <= L + mid - 1 && a[pos[p[j]]] <= L + mid - 1)
          x = nxtR[i], y = preR[j],
          (x <= y ? res[p[i]][p[j]] = Merge(tmp[p[i]][p[j]], tmp[p[x]][p[y]])
                  : 0);
        else if (a[pos[p[i]]] > L + mid - 1 && a[pos[p[j]]] > L + mid - 1)
          x = nxtL[i], y = preL[j],
          (x <= y ? res[p[i]][p[j]] = Merge(tmp[p[x]][p[y]], tmp[p[i]][p[j]])
                  : 0);
        else if (a[pos[p[i]]] <= L + mid - 1 && a[pos[p[j]]] > L + mid - 1)
          x = nxtR[i], y = preL[j],
          res[p[i]][p[j]] = Merge(tmp[p[i]][p[y]], tmp[p[x]][p[j]]);
        else if (a[pos[p[i]]] > L + mid - 1 && a[pos[p[j]]] <= L + mid - 1)
          x = nxtL[i], y = preR[j],
          res[p[i]][p[j]] = Merge(tmp[p[x]][p[j]], tmp[p[i]][p[y]]);
      }
  }
  inline void build(int l, int r) {
    L = l, R = r;
    for (int i = (1); i <= (n); i++)
      if (a[i] >= l && a[i] <= r) pos[++N] = i, rev[a[i]] = N;
    vector<int> p(0);
    for (int i = (1); i <= (N); i++) p.push_back(i);
    solve(1, N, p);
  }
  inline int ask(int l, int r) {
    int L = lower_bound(pos + 1, pos + N + 1, l) - pos,
        R = upper_bound(pos + 1, pos + N + 1, r) - pos - 1;
    if (L > R) return 0;
    return res[L][R];
  }
} p[Sz];
int res[MAXM];
int main() {
  n = read(), m = read(), cnt = n;
  for (int i = (1); i <= (n); i++) a[i] = read();
  for (int i = (1); i <= ((n - 1) / B + 1); i++)
    p[i].build((i - 1) * B + 1, min(i * B, n));
  for (int i = (1); i <= (m); i++) {
    int l = read(), r = read();
    res[i] = 0;
    for (int j = (1); j <= ((n - 1) / B + 1); j++) {
      int s = p[j].ask(l, r);
      if (!s) continue;
      if (!res[i])
        res[i] = s;
      else
        res[i] = Merge(res[i], s);
    }
  }
  cout << cnt << "\n";
  for (int i = (n + 1); i <= (cnt); i++)
    cout << ans[i].first << " " << ans[i].second << "\n";
  for (int i = (1); i <= (m); i++) cout << res[i] << ' ';
  cout << "\n";
  return 0;
}