题意：

$n$ 长度序列上 $q$ 次区间 $+x$ , 查询区间历史 $\geq c$ 时长。

题解：

序列和时间二维平面上的矩形加柱形 $\geq c$ 查询，扫描线转化成区间加区间 $\geq c$ 查询。

#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define blk(x) ((x) / bs)
#define hob(x) ((x)*bs)
#define eob(x) min(((x)+1) * bs - 1, n)
using namespace std;

const int maxn = 1e5, bs = 316, nb = maxn / bs + 10;
struct block {
    ll a[maxn + 10], tag[nb + 10];
    int ovi[maxn + 10], buf1[bs + 10], buf2[bs + 10], n;
    void init(int _n) {
        n = _n;
        rep(i,0,n) ovi[i] = i;
    }
    void add(int x, int c) {
        int bx = blk(x);
        rep(i,bx + 1, blk(n)) tag[i] += c;
        int s = hob(bx), t = eob(bx);
        rep(i,x,t) a[i] += c;
        int e1 = -1, e2 = -1, p1 = 0, p2 = 0;
        rep(i,s,t) if(ovi[i] >= x) buf1[++ e1] = ovi[i]; else buf2[++ e2] = ovi[i];
        rep(i,s,t) ovi[i] = (p2 > e2 || p1 <= e1 && a[buf1[p1]] < a[buf2[p2]]) ? buf1[p1 ++] : buf2[p2 ++];
    }
    int qry(int x, int c) {
        int bx = blk(x), ans = 0;
        rep(i,blk(0),bx - 1) {
            int s = hob(i), t = eob(i);
            if(a[ovi[t]] + tag[i] < c) continue;
            if(a[ovi[s]] + tag[i] >= c) { ans += t - s + 1; continue; }
            while(s < t) {
                int mid = (s + t) >> 1;
                if(a[ovi[mid]] + tag[i] >= c) t = mid;
                else s = mid + 1;
            }
            ans += eob(i) - s + 1;
        }
        rep(i,hob(bx),x) ans += a[i] + tag[bx] >= c;
        return ans;
    }
} blker;

struct node { int x, c; };
vector< node > v[maxn + 10];
vector< pair<node, int> > q[maxn + 10];

int ans[maxn + 10], a[maxn + 10];

int main() {
    int n, totQ, opt, l, r, x, p, y;
    scanf("%d %d", &n, &totQ);
    rep(i,1,n) scanf("%d", a + i);
    blker.init(totQ);
    memset(ans, -1, sizeof(ans));
    rep(qr,1,totQ) {
        scanf("%d", &opt);
        if(opt == 1) {
            scanf("%d %d %d", &l, &r, &x);
            v[l].push_back({qr, x});
            if(r != n) v[r + 1].push_back({qr, -x});
        } else if(opt == 2) {
            scanf("%d %d", &p, &y);
            q[p].push_back({{qr - 1, y}, qr});
        }
    }
    rep(i,1,n) {
        for(auto &it : v[i]) {
            blker.add(it.x, it.c);
        }
        for(auto &it : q[i]) {
            ans[it.second] = blker.qry(it.first.x, it.first.c - a[i]);
        }
    }
    rep(i,1,totQ) if(~ans[i]) printf("%d\n", ans[i]);
    return 0;
}