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;
const long double pi = acos(-1);
const int n_ = 2e5 + 5;
int n, q;
long long st[n_ << 4], lazy[n_ << 4];
set<pair<pair<int, int>, int>> r;
void upd(long long v, int s, int e, int id = 1, int l = 1, int r = n) {
  int mid = (l + r) >> 1;
  if (lazy[id]) {
    lazy[id << 1] += lazy[id];
    lazy[id << 1 | 1] += lazy[id];
    st[id << 1] += lazy[id] * (mid - l + 1);
    st[id << 1 | 1] += lazy[id] * (r - mid);
    lazy[id] = 0;
  }
  if (s > r || e < l) return;
  if (s <= l && r <= e) {
    lazy[id] = v;
    st[id] += v * (r - l + 1);
    return;
  }
  upd(v, s, e, id << 1, l, mid);
  upd(v, s, e, id << 1 | 1, mid + 1, r);
  st[id] = st[id << 1] + st[id << 1 | 1];
}
long long query(int s, int e, int id = 1, int l = 1, int r = n) {
  int mid = (l + r) >> 1;
  if (lazy[id]) {
    lazy[id << 1] += lazy[id];
    lazy[id << 1 | 1] += lazy[id];
    st[id << 1] += lazy[id] * (mid - l + 1);
    st[id << 1 | 1] += lazy[id] * (r - mid);
    lazy[id] = 0;
  }
  if (s > r || e < l) return 0;
  if (s <= l && r <= e) return st[id];
  return query(s, e, id << 1, l, mid) + query(s, e, id << 1 | 1, mid + 1, r);
}
void make(int &s, int &e, int &x) {
  pair<pair<int, int>, int> cur;
  int nl, nr, nnl, nnr, ns = s;
  while (ns <= e) {
    cur = *(r.lower_bound({{ns, 0}, 0}));
    r.erase(cur);
    nl = cur.first.second;
    nr = ns - 1;
    if (nl <= nr) r.insert({{nr, nl}, cur.second});
    if (e < cur.first.first) {
      nnl = e + 1;
      nnr = cur.first.first;
      upd(abs(x - cur.second), ns, e);
      r.insert({{nnr, nnl}, cur.second});
      break;
    }
    upd(abs(x - cur.second), ns, cur.first.first);
    ns = cur.first.first + 1;
  }
  r.insert({{e, s}, x});
}
int main() {
  scanf("%d %d", &n, &q);
  int t, s, e, x;
  for (int i = 1; i <= n; i++) r.insert({{i, i}, i});
  while (q--) {
    scanf("%d %d %d", &t, &s, &e);
    if (t & 1) {
      scanf("%d", &x);
      make(s, e, x);
    } else {
      printf("%lld\n", query(s, e));
    }
  }
  return 0;
}
