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 std::cin;
using std::cout;
const int N = 4100, BL = 8, B = 1 << BL;
int C2[N];
inline int merge(int, int);
struct block {
  std::vector<int> v, f;
  block() {}
  block(const std::vector<int> &val) {
    const int n = val.size();
    assert(n);
    if (n == 1) {
      v.emplace_back(val.back()), f.emplace_back(val.back());
      return;
    }
    std::vector<int>::const_iterator middle = val.begin() + n / 2;
    *this = block(std::vector<int>(val.begin(), middle)) +
            block(std::vector<int>(middle, val.end()));
  }
  inline int operator()(int L, int R) const {
    L = std::lower_bound(v.begin(), v.end(), L) - v.begin();
    R = std::upper_bound(v.begin(), v.end(), R) - v.begin() - 1;
    return L <= R ? f[L + C2[R]] : 0;
  }
  friend inline block operator+(const block &L, const block &R) {
    int i, j, Ln = L.v.size(), Rn = R.v.size(), n = Ln + Rn;
    block ret;
    ret.v.resize(n), ret.f.clear(), ret.f.reserve(C2[n + 1]);
    std::merge(L.v.begin(), L.v.end(), R.v.begin(), R.v.end(), ret.v.begin());
    for (j = 0; j < n; ++j)
      for (i = 0; i <= j; ++i)
        ret.f.emplace_back(merge(L(ret.v[i], ret.v[j]), R(ret.v[i], ret.v[j])));
    return ret;
  }
};
int n, m, q, O;
int p[N], ans[66000];
block s[B];
std::pair<int, int> op[2200000];
inline int min(const int x, const int y) { return x < y ? x : y; }
inline int merge(int A, int B) {
  return A && B ? (op[O] = std::pair<int, int>(A, B), ++O) : A | B;
}
int main() {
  int i, j, L, R, len;
  std::vector<int> t, u;
  std::ios::sync_with_stdio(false), cin.tie(NULL);
  for (i = 1; i < N; ++i) C2[i] = C2[i - 1] + i;
  cin >> n >> q, O = n, m = ((n - 1) >> BL) + 1;
  for (i = 1; i <= n; ++i) cin >> j, p[--j] = i;
  for (i = 0; i < m; ++i) {
    L = i << BL, R = min((i + 1) << BL, n), t.clear(), t.reserve(len = R - L);
    for (j = L; j < R; ++j) t.emplace_back(p[j]);
    s[i] = block(t);
  }
  for (i = 0; i < q; ++i) {
    cin >> L >> R;
    for (j = 0; j < m; ++j) ans[i] = merge(ans[i], s[j](L, R));
  }
  cout << O << '\n';
  for (i = n; i < O; ++i) cout << op[i].first << ' ' << op[i].second << '\n';
  for (i = 0; i < q; ++i) cout << ans[i] << (i == q - 1 ? '\n' : ' ');
  return 0;
}