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;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string& s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
int n;
vector<int> a;
vector<pair<int, int>> gen;
vector<int> Solve(int L, int R, const vector<pair<int, int>>& qs) {
  int q = (int)qs.size();
  vector<int> ret(q, -1);
  if (L == R) {
    int pos = -1;
    for (int i = 0; i < n; i++) {
      if (a[i] == L) {
        pos = i;
      }
    }
    assert(pos != -1);
    for (int i = 0; i < q; i++) {
      ret[i] = pos;
    }
  } else {
    int M = (L + R) >> 1;
    vector<int> next_lower(n, n);
    vector<int> next_upper(n, n);
    vector<int> prev_lower(n, -1);
    vector<int> prev_upper(n, -1);
    for (int i = 0; i < n; i++) {
      if (L <= a[i] && a[i] <= M) {
        prev_lower[i] = i;
      } else {
        prev_lower[i] = (i == 0 ? -1 : prev_lower[i - 1]);
      }
      if (M < a[i] && a[i] <= R) {
        prev_upper[i] = i;
      } else {
        prev_upper[i] = (i == 0 ? -1 : prev_upper[i - 1]);
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      if (L <= a[i] && a[i] <= M) {
        next_lower[i] = i;
      } else {
        next_lower[i] = (i == n - 1 ? n : next_lower[i + 1]);
      }
      if (M < a[i] && a[i] <= R) {
        next_upper[i] = i;
      } else {
        next_upper[i] = (i == n - 1 ? n : next_upper[i + 1]);
      }
    }
    vector<int> id_lower(q, -1);
    vector<int> id_upper(q, -1);
    {
      vector<pair<int, int>> qs_lower;
      for (int i = 0; i < q; i++) {
        int x = next_lower[qs[i].first];
        int y = prev_lower[qs[i].second];
        if (x <= y) {
          qs_lower.emplace_back(x, y);
        }
      }
      sort(qs_lower.begin(), qs_lower.end());
      qs_lower.resize(unique(qs_lower.begin(), qs_lower.end()) -
                      qs_lower.begin());
      auto got_lower = Solve(L, M, qs_lower);
      for (int i = 0; i < q; i++) {
        int x = next_lower[qs[i].first];
        int y = prev_lower[qs[i].second];
        if (x <= y) {
          auto it =
              lower_bound(qs_lower.begin(), qs_lower.end(), make_pair(x, y));
          assert(it != qs_lower.end() && it->first == x && it->second == y);
          id_lower[i] = got_lower[(int)(it - qs_lower.begin())];
        }
      }
    }
    {
      vector<pair<int, int>> qs_upper;
      for (int i = 0; i < q; i++) {
        int x = next_upper[qs[i].first];
        int y = prev_upper[qs[i].second];
        if (x <= y) {
          qs_upper.emplace_back(x, y);
        }
      }
      sort(qs_upper.begin(), qs_upper.end());
      qs_upper.resize(unique(qs_upper.begin(), qs_upper.end()) -
                      qs_upper.begin());
      auto got_upper = Solve(M + 1, R, qs_upper);
      for (int i = 0; i < q; i++) {
        int x = next_upper[qs[i].first];
        int y = prev_upper[qs[i].second];
        if (x <= y) {
          auto it =
              lower_bound(qs_upper.begin(), qs_upper.end(), make_pair(x, y));
          assert(it != qs_upper.end() && it->first == x && it->second == y);
          id_upper[i] = got_upper[(int)(it - qs_upper.begin())];
        }
      }
    }
    for (int i = 0; i < q; i++) {
      assert(id_lower[i] != -1 || id_upper[i] != -1);
      if (id_lower[i] == -1) {
        ret[i] = id_upper[i];
        continue;
      }
      if (id_upper[i] == -1) {
        ret[i] = id_lower[i];
        continue;
      }
      ret[i] = (int)gen.size();
      gen.emplace_back(id_lower[i], id_upper[i]);
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> n >> q;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    --a[i];
  }
  vector<pair<int, int>> qs(q);
  for (int i = 0; i < q; i++) {
    cin >> qs[i].first >> qs[i].second;
    --qs[i].first;
    --qs[i].second;
  }
  gen.assign(n, make_pair(-1, -1));
  auto res = Solve(0, n - 1, qs);
  cout << (int)gen.size() << '\n';
  for (int i = n; i < (int)gen.size(); i++) {
    cout << gen[i].first + 1 << " " << gen[i].second + 1 << '\n';
  }
  for (int i = 0; i < (int)res.size(); i++) {
    if (i > 0) {
      cout << " ";
    }
    cout << res[i] + 1;
  }
  cout << '\n';
  return 0;
}