Problem: Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number y, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to  - 1. GukiZ also prepared a problem for you. This time, you have two types of queries: 

  1. First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x. 
  2. Second type has form 2 y and asks you to find value of GukiZiana(a, y). 



For each query of type 2, print the answer and make GukiZ happy!

Input

The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries. 

The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a. 

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.

Output

For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

Examples

Input

4 3
1 2 3 4
1 1 2 1
1 1 1 1
2 3


Output

2


Input

2 3
1 2
1 2 2 1
2 3
2 4


Output

0
-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
std::istream& operator>>(std::istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T, typename U>
std::ostream& operator<<(std::ostream& o, const pair<T, U>& p) {
  o << p.first << ' ' << p.second;
  return o;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  if (t.empty()) o << '\n';
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
bool in(T a, T b, T c) {
  return a <= b && b < c;
}
unsigned int logceil(int first) {
  return 8 * sizeof(int) - __builtin_clz(first);
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.first) ^ (u(p.second) << 7);
  }
};
}  // namespace std
template <typename T, typename F>
T bsh(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      l = m + 1;
      r = m;
    } else {
      h = m - 1;
    }
  }
  return r;
}
template <typename F>
double bshd(double l, double h, const F& f, double p = 1e-9) {
  unsigned int r = 3 + (unsigned int)log2((h - l) / p);
  while (r--) {
    double m = (l + h) / 2;
    if (f(m)) {
      l = m;
    } else {
      h = m;
    }
  }
  return (l + h) / 2;
}
template <typename T, typename F>
T bsl(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      h = m - 1;
      r = m;
    } else {
      l = m + 1;
    }
  }
  return r;
}
template <typename F>
double bsld(double l, double h, const F& f, double p = 1e-9) {
  unsigned int r = 3 + (unsigned int)log2((h - l) / p);
  while (r--) {
    double m = (l + h) / 2;
    if (f(m)) {
      h = m;
    } else {
      l = m;
    }
  }
  return (l + h) / 2;
}
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector2<T>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector2<T>>(a, vector2<T>(b, c, t)) {}
};
template <typename T>
class vector4 : public vector<vector3<T>> {
 public:
  vector4() {}
  vector4(size_t a, size_t b, size_t c, size_t d, T t = T())
      : vector<vector3<T>>(a, vector3<T>(b, c, d, t)) {}
};
template <typename T>
class vector5 : public vector<vector4<T>> {
 public:
  vector5() {}
  vector5(size_t a, size_t b, size_t c, size_t d, size_t e, T t = T())
      : vector<vector4<T>>(a, vector4<T>(b, c, d, e, t)) {}
};
constexpr int inf = 1e9 + 1;
class gukizandgukizuana {
 public:
  struct Strip {
    int L, R;
    int X;
    vector<int> A;
    unordered_map<int, int> Left;
    unordered_map<int, int> Right;
    Strip(const vector<int>& X, int l, int r) : L(l), R(r), X(0), A(R - L) {
      for (int i = 0; i < A.size(); ++i) A[i] = X[i + L];
      recalc();
    }
    void recalc() {
      Left.clear();
      Right.clear();
      for (int i = 0; i < A.size(); ++i) Right[A[i]] = i + L;
      for (int i = (int)A.size() - 1; i >= 0; --i) Left[A[i]] = i + L;
    }
    void increase(int l, int r, int first) {
      if (X == inf)
        return;
      else if (l <= L && r >= R)
        X = min(inf, X + first);
      else if (l >= R || r <= L)
        return;
      else {
        if (X)
          for (int i = 0; i < A.size(); ++i) A[i] = min(inf, A[i] + X);
        for (int i = max(l, L) - L; i < min(r, R) - L; ++i)
          A[i] = min(inf, A[i] + first);
        X = 0;
        recalc();
      }
    }
    int left(int second) const {
      if (X == inf) return inf;
      auto it = Left.find(second - X);
      return it != Left.end() ? it->second : inf;
    }
    int right(int second) const {
      if (X == inf) return -1;
      auto it = Right.find(second - X);
      return it != Right.end() ? it->second : -1;
    }
  };
  void solve(istream& cin, ostream& cout) {
    int N, M;
    cin >> N >> M;
    vector<int> A(N);
    cin >> A;
    constexpr int BLOCK = 600;
    vector<Strip> S;
    for (int i = 0; i < N; i += BLOCK) S.emplace_back(A, i, min(i + BLOCK, N));
    for (int i = 0; i < M; ++i) {
      int t;
      cin >> t;
      if (t == 1) {
        int l, r, first;
        cin >> l >> r >> first;
        for (auto& s : S) {
          s.increase(l - 1, r, first);
        }
      } else {
        int l = inf, r = -1;
        int second;
        cin >> second;
        for (auto& s : S) {
          l = min(l, s.left(second));
          r = max(r, s.right(second));
        }
        cout << (r == -1 ? -1 : r - l) << '\n';
      }
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  gukizandgukizuana solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}