DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

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


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
inline long long myAbs(long long x) { return x > 0 ? x : -x; }
int n, m;
class st {
 public:
  vector<long long> c, lazy;
  st() : c(4 * n + 1, 0), lazy(4 * n + 1, 0) {}
  void update(int l, int r, long long v, int nl, int nr, int node) {
    if (l > nr || r < nl) return;
    if (l <= nl && nr <= r) {
      lazy[node] += v;
      c[node] += v * (nr - nl + 1);
      return;
    }
    update(l, r, v, nl, (nl + nr) / 2, node * 2);
    update(l, r, v, (nl + nr) / 2 + 1, nr, node * 2 + 1);
    c[node] = c[node * 2] + c[node * 2 + 1] + lazy[node] * (nr - nl + 1);
  }
  void update(int l, int r, long long v) { update(l, r, v, 1, n, 1); }
  long long query(int l, int r, int nl, int nr, int node) {
    if (l > nr || r < nl) return 0;
    if (l <= nl && nr <= r) {
      return c[node];
    }
    return query(l, r, nl, (nl + nr) / 2, node * 2) +
           query(l, r, (nl + nr) / 2 + 1, nr, node * 2 + 1) +
           lazy[node] * (min(r, nr) - max(l, nl) + 1);
  }
  long long query(int l, int r) { return query(l, r, 1, n, 1); }
};
class painting {
 public:
  st tree;
  set<pair<pair<int, int>, int> > s;
  painting(int n) {
    s = set<pair<pair<int, int>, int> >();
    for (int i = 1; i <= n; i++)
      s.insert(pair<pair<int, int>, int>(pair<int, int>(i, i), i));
  }
  void update(int l, int r, int x) {
    set<pair<pair<int, int>, int> >::iterator lb, rb;
    auto _lb =
        s.lower_bound(pair<pair<int, int>, int>(pair<int, int>(l, l), 0));
    if (_lb->first.first != l) {
      _lb--;
      pair<pair<int, int>, int> that = *_lb;
      s.erase(_lb);
      s.insert(pair<pair<int, int>, int>(
          pair<int, int>(that.first.first, l - 1), that.second));
      s.insert(pair<pair<int, int>, int>(pair<int, int>(l, that.first.second),
                                         that.second));
    }
    auto _rb = s.lower_bound(
        pair<pair<int, int>, int>(pair<int, int>(r + 1, r + 1), 0));
    if (_rb == s.end() || _rb->first.first != r + 1) {
      _rb--;
      pair<pair<int, int>, int> that = *_rb;
      s.erase(_rb);
      s.insert(pair<pair<int, int>, int>(pair<int, int>(that.first.first, r),
                                         that.second));
      s.insert(pair<pair<int, int>, int>(
          pair<int, int>(r + 1, that.first.second), that.second));
    }
    lb = s.lower_bound(pair<pair<int, int>, int>(pair<int, int>(l, l), 0));
    rb = s.lower_bound(
        pair<pair<int, int>, int>(pair<int, int>(r + 1, r + 1), 0));
    for (auto it = lb; it != rb; it++)
      tree.update(it->first.first, it->first.second, myAbs(it->second - x));
    s.erase(lb, rb);
    s.insert(pair<pair<int, int>, int>(pair<int, int>(l, r), x));
  }
  long long query(int l, int r) { return tree.query(l, r); }
};
int main() {
  cin >> n >> m;
  painting p = painting(n);
  while (m--) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r, x;
      cin >> l >> r >> x;
      p.update(l, r, x);
    } else {
      int l, r;
      cin >> l >> r;
      cout << p.query(l, r) << endl;
    }
  }
}
