description: You are given two integer arrays of length N, A1 and A2. You are also given Q queries of 4 types: 
  
  1 k l r x: set Ak_i:=min(Ak_i, x) for each l ≤ i ≤ r.
  
  2 k l r x: set Ak_i:=max(Ak_i, x) for each l ≤ i ≤ r.
  
  3 k l r x: set Ak_i:=Ak_i+x for each l ≤ i ≤ r.
  
  4 l r: find the (∑_{i=l}^r F(A1_i+A2_i)) \% (10^9+7) where F(k) is the k-th Fibonacci number (F(0)=0, F(1)=1, F(k)=F(k-1)+F(k-2)), and x \% y denotes the remainder of the division of x by y.
  
  You should process these queries and answer each query of the fourth type.
  
  Input
  
  The first line contains two integers N and Q. (1 ≤ N, Q ≤ 5 × 10^4) 
  
  The second line contains N integers, array A1_1, A1_2, ... A1_N. (0 ≤ A1_i ≤ 10^6)
  
  The third line contains N integers, array A2_1, A2_2, ... A2_N. (0 ≤ A2_i ≤ 10^6)
  
  The next Q lines describe the queries. Each line contains 5 or 3 integers, where the first integer denotes the type of the query. (k ∈ \{1, 2\}, 1 ≤ l ≤ r ≤ N)
  
  For queries of type 1 and 2, 0 ≤ x ≤ 10^9 holds.
  
  For queries of type 3, −10^6 ≤ x ≤ 10^6 holds.
  
  It is guaranteed that after every query each number in arrays A1 and A2 will be nonnegative. 
  
  Output
  
  Print the answer to each query of the fourth type, in separate lines. 
  
  Examples
  
  Input
  
  
  3 4
  1 0 2
  2 1 0
  4 1 3
  3 2 2 2 3
  1 1 1 3 0
  4 1 3
  
  
  Output
  
  
  4
  4
  
  
  Input
  
  
  5 4
  1 3 5 3 2
  4 2 1 3 3
  4 1 3
  4 2 5
  2 1 2 4 6
  4 2 4
  
  
  Output
  
  
  18
  26
  68
  
  Note
  
  In the first example: The answer for the first query is F(1 + 2) + F(0 + 1) + F(2 + 0) = F(3) + F(1) + F(2) = 2 + 1 + 1 = 4. After the second query, the array A2 changes to [2, 4, 0]. After the third query, the array A1 changes to [0, 0, 0]. The answer for the fourth query is F(0 + 2) + F(0 + 4) + F(0 + 0) = F(2) + F(4) + F(0) = 1 + 3 + 0 = 4. 
  
  In the second example: The answer for the first query is F(1 + 4) + F(3 + 2) + F(5 + 1) = F(5) + F(5) + F(6) = 5 + 5 + 8 = 18. The answer for the second query is F(3 + 2) + F(5 + 1) + F(3 + 3) + F(2 + 3) = F(5) + F(6) + F(6) + F(5) = 5 + 8 + 8 + 5 = 26. After the third query, the array A1 changes to [1, 6, 6, 6, 2]. The answer for the fourth query is F(6 + 2) + F(6 + 1) + F(6 + 3) = F(8) + F(7) + F(9) = 21 + 13 + 34 = 68. 
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long MOD = 1e9 + 7;
  struct Fib {
    long long a, b;
    Fib operator+(const Fib& o) const {
      return {(a + o.a) % MOD, (b + o.b) % MOD};
    }
  };
  struct FibOp {
    long long a, b, c, d;
    FibOp operator()(const FibOp& o) const {
      FibOp res = {a * o.a + b * o.c, a * o.b + b * o.d, c * o.a + d * o.c,
                   c * o.b + d * o.d};
      res.a %= MOD;
      ;
      res.b %= MOD;
      ;
      res.c %= MOD;
      ;
      res.d %= MOD;
      ;
      return res;
    }
    Fib operator()(const Fib& f) const {
      Fib res = {a * f.a + b * f.b, c * f.a + d * f.b};
      res.a %= MOD;
      ;
      res.b %= MOD;
      ;
      return res;
    }
  };
  const long long K = 32;
  FibOp pwr[K];
  FibOp negpwr[K];
  Fib zero = {0, 1};
  void setup() {
    pwr[0] = {1, 1, 1, 0};
    negpwr[0] = {0, 1, 1, MOD - 1};
    for (int i = 1; i < K; ++i) {
      pwr[i] = pwr[i - 1](pwr[i - 1]);
      negpwr[i] = negpwr[i - 1](negpwr[i - 1]);
    }
  }
  Fib FibMult(Fib f, long long cnt, FibOp p[K]) {
    for (int r = K - 1; r >= 0; --r) {
      if (cnt & (1LL << r)) {
        f = p[r](f);
      }
    }
    return f;
  }
  Fib FibMult(Fib f, long long cnt) {
    return (cnt >= 0 ? FibMult(f, cnt, pwr) : FibMult(f, -cnt, negpwr));
  }
  struct Node {
    long long lo[2] = {0, 0};
    long long hi[2] = {0, 0};
    long long add[2] = {0, 0};
    long long sz = 1;
    Fib f;
    void apply(long long a[2]) {
      for (int i = 0; i < 2; ++i) {
        add[i] += a[i];
        lo[i] += a[i];
        hi[i] += a[i];
      }
      f = FibMult(f, a[0] + a[1]);
    }
  };
  struct Segtree {
    vector<Node> t;
    int n, h;
    Segtree(int sz) {
      h = 32 - __builtin_clz(sz);
      n = 1 << h;
      t = vector<Node>(n << 1);
      for (int i = n - 1; i > 0; --i) {
        t[i].sz = t[i * 2].sz + t[i * 2 + 1].sz;
      }
    }
    void push(int x) {
      if (!t[x].add[0] && !t[x].add[1]) {
        return;
      }
      assert(x < n);
      t[x << 1].apply(t[x].add);
      t[x << 1 | 1].apply(t[x].add);
      t[x].add[0] = 0;
      t[x].add[1] = 0;
    }
    void pull(int x) {
      for (int i = 0; i < 2; ++i) {
        t[x].lo[i] = min(t[x * 2].lo[i], t[x * 2 + 1].lo[i]);
        t[x].hi[i] = max(t[x * 2].hi[i], t[x * 2 + 1].hi[i]);
      }
      t[x].f = t[x << 1].f + t[x << 1 | 1].f;
      t[x].apply(t[x].add);
    }
    void set_init(int x, int k, int va) {
      x += n;
      t[x].lo[k] = t[x].hi[k] = va;
      t[x].f = FibMult(zero, t[x].lo[0] + t[x].lo[1]);
      for (x /= 2; x; x /= 2) {
        pull(x);
      }
    }
    long long query(int l, int r, int x, int lv, int rv) {
      if (r <= lv || rv <= l) {
        return 0;
      }
      if (l <= lv && rv <= r) {
        return t[x].f.a;
      }
      push(x);
      int mv = (lv + rv) / 2;
      long long ans = 0;
      ans += query(l, r, x * 2, lv, mv);
      ans += query(l, r, x * 2 + 1, mv, rv);
      return ans % MOD;
    }
    long long query(int l, int r) { return query(l, r, 1, 0, n); }
    void add(int l, int r, int k, long long va, int x, int lv, int rv) {
      if (r <= lv || rv <= l) {
        return;
      }
      if (l <= lv && rv <= r) {
        long long a[2] = {0, 0};
        a[k] = va;
        t[x].apply(a);
        return;
      }
      push(x);
      int mv = (lv + rv) / 2;
      add(l, r, k, va, x * 2, lv, mv);
      add(l, r, k, va, x * 2 + 1, mv, rv);
      pull(x);
    }
    void add(int l, int r, int k, long long va) { add(l, r, k, va, 1, 0, n); }
    void minupdate(int l, int r, int k, long long va, int x, int lv, int rv) {
      if (r <= lv || rv <= l) {
        return;
      }
      if (t[x].hi[k] <= va) {
        return;
      }
      if (l <= lv && rv <= r && t[x].lo[k] == t[x].hi[k]) {
        long long a[2] = {0, 0};
        a[k] = va - t[x].lo[k];
        t[x].apply(a);
        return;
      }
      push(x);
      int mv = (lv + rv) / 2;
      minupdate(l, r, k, va, x * 2, lv, mv);
      minupdate(l, r, k, va, x * 2 + 1, mv, rv);
      pull(x);
    }
    void minupdate(int l, int r, int k, long long va) {
      return minupdate(l, r, k, va, 1, 0, n);
    }
    void maxupdate(int l, int r, int k, long long va, int x, int lv, int rv) {
      if (r <= lv || rv <= l) {
        return;
      }
      if (t[x].lo[k] >= va) {
        return;
      }
      if (l <= lv && rv <= r && t[x].lo[k] == t[x].hi[k]) {
        long long a[2] = {0, 0};
        a[k] = va - t[x].lo[k];
        t[x].apply(a);
        return;
      }
      push(x);
      int mv = (lv + rv) / 2;
      maxupdate(l, r, k, va, x * 2, lv, mv);
      maxupdate(l, r, k, va, x * 2 + 1, mv, rv);
      pull(x);
    }
    void maxupdate(int l, int r, int k, long long va) {
      return maxupdate(l, r, k, va, 1, 0, n);
    }
  };
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    setup();
    int n, q;
    cin >> n >> q;
    Segtree st(n + 10);
    for (int i = 1; i <= n; ++i) {
      int x;
      cin >> x;
      st.set_init(i, 0, x);
    }
    for (int i = 1; i <= n; ++i) {
      int x;
      cin >> x;
      st.set_init(i, 1, x);
    }
    for (int _i = 0; _i < q; ++_i) {
      int type;
      cin >> type;
      if (type == 1) {
        int k, l, r, x;
        cin >> k >> l >> r >> x;
        --k;
        st.minupdate(l, r + 1, k, x);
      } else if (type == 2) {
        int k, l, r, x;
        cin >> k >> l >> r >> x;
        --k;
        st.maxupdate(l, r + 1, k, x);
      } else if (type == 3) {
        int k, l, r, x;
        cin >> k >> l >> r >> x;
        --k;
        st.add(l, r + 1, k, x);
      } else if (type == 4) {
        int l, r;
        cin >> l >> r;
        cout << st.query(l, r + 1) << "\n";
      }
    }
    return 0;
  }
  